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2 years ago

Among timber, wind, and oil, which is a fossil fuel and which is not?

Chemistry

2 answers:

Cloud [144]2 years ago

5 0

Answer:

Oil is a fossil fuel and timber or wood and wind is not. Hope this helps!

Rus_ich [418]2 years ago

5 0

Answer:

Ello Lad!

Wind and Timber are not fossil fuels, these are renewable resources as well! Oil has something called hydrocarbons therefore making it a fossil fuel!

Hope this helped :)

have a good day

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Under which of the following conditions would a chemical reaction most likely proceed at the fastest rate?

Valentin [98]

<h2>Answer:</h2>

Option A is correct

Adding an enzyme to decrease the activation energy of the reaction

<h2>Explanation:</h2>

Enzymes are the biological catalyst. They are proteins in nature. They are naturally found in humans,animals,micro-organisms,plants etc. They catalyze the chemical reactions by lowering activation energy and without being consumed in it.

6 0

3 years ago

Read 2 more answers

Calculate the mass of 25,000 molecules of nitrogen gas. (1 mole = 6.02 x 1023 molecules)

Ainat [17]

Hey there!

Molar mass N2 = 28.01 g/mol

Therefore:

28.01 g N2 -------------- 6.02*10²² molecules N2

( mass N2 ?? ) ----------- 25,000 molecules N2

mass N2 = ( 25,000 * 28.01 ) / ( 6.02*10²³ )

mass N2 = 700250 / 6.02*10²³

mass N2 = 1.163*10⁻¹⁸ g

Hope that helps!

7 0

3 years ago

Read 2 more answers

Be sure to answer all parts. Consider the formation of ammonia in two experiments. (a) To a 1.00−L container at 727°C, 1.30 mol

Sonbull [250]

<u>Answer:</u> The value of $K_c$ for $2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$ reaction is $5.13\times 10^2$

<u>Explanation:</u>

We are given:

Initial moles of nitrogen gas = 1.30 moles

Initial moles of hydrogen gas = 1.65 moles

Equilibrium moles of ammonia = 0.100 moles

Volume of the container = 1.00 L

For the given chemical equation:

$N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)$

<u>Initial:</u> 1.30 1.65

<u>At eqllm:</u> 1.30-x 1.65-3x 2x

Evaluating the value of 'x'

$\Rightarrow 2x=0.100\\\\\Rightarrow x=0.050mol$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[NH_3]^2}{[N_2]\times [H_2]^3}$

Equilibrium moles of nitrogen gas = $(1.30-x)=(1.30-0.05)=1.25mol$

Equilibrium moles of hydrogen gas = $(1.65-x)=(1.65-0.05)=1.60mol$

Putting values in above expression, we get:

$K_c=\frac{(0.100)^2}{1.25\times (1.60)^3}\\\\K_c=1.95\times 10^{-3}$

Calculating the $K_c'$ for the given chemical equation:

$2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$

$K_c'=\frac{1}{K_c}\\\\K_c'=\frac{1}{1.95\times 10^{-3}}=5.13\times 10^2$

Hence, the value of $K_c$ for $2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$ reaction is $5.13\times 10^2$

8 0

3 years ago

Read 2 more answers

PLEASE HELP!! GIVING BRAINLIEST

RSB [31]

Answer:

The answer is "0.00172172603".

Explanation:

Given:

$Mass (M) = 1.60 \times 10^{-3} \ g\\\\Density (D) = 9.293 \times 10^{-1} \ \frac{g}{cm^3}\\\\Volume (V) = ?$

Formula:

$\to \bold{V = \frac{M}{D}}$

$= \frac{1.60 \times 10^{-3}}{9.293 \times 10^{-1}} \\\\ = \frac{1.60 \times 10^{1}}{9.293 \times 10^{3}} \\\\ = \frac{1.60 \times 10}{9.293 \times 1000} \\\\ = \frac{1.60 }{9.293 \times 100} \\\\ = \frac{1.60 }{929.3 } \\\\= 0.00172172603$

4 0

3 years ago

What is the preasure in atmospheres of 20 mol of nitrogen gas in 36.2 L cylinder at 25 degrees C?

Savatey [412]

Answer:

P = 13.5 atm

Explanation:

Given that

No. of moles, n = 20 moles

Volume of nitrogen gas = 36.2 L

Temperature = 25°C = 298 K

We need to find the pressure of the gas. Using the ideal gas equation

PV = nRT

Where

R is gas constant, $R=0.082057\ L-atm/K-mol$

So,

$P=\dfrac{nRT}{V}\\\\P=\dfrac{20\times 0.082057\times 298}{36.2 }\\\\P=13.5\ atm$

so, the pressure of the gas is equal to 13.5 atm.

7 0

2 years ago